Core Problems for
AP Calculus BC

This is 0/0 indeterminate form — apply L'Hôpital's Rule (differentiate top and bottom separately).

• Numerator derivative: \(\dfrac{d}{dx}[\tan(3x)] = 3\sec^2(3x)\)
• Denominator derivative: \(\dfrac{d}{dx}[5x] = 5\)
• Evaluate at \(x=0\): \(\dfrac{3\sec^2(0)}{5} = \dfrac{3 \cdot 1}{5} = \dfrac{3}{5}\)

Key insight: Any limit of the form \(\lim_{x\to 0}\dfrac{\tan(kx)}{mx} = \dfrac{k}{m}\).

Set left-hand limit = right-hand limit at \(x=3\):

• Left: \(\lim_{x \to 3^-}(x^2+2) = 9+2 = 11\)
• Right: \(f(3) = k(3)-1 = 3k-1\)
• Set equal: \(3k - 1 = 11 \Rightarrow 3k = 12 \Rightarrow k = 4\)

Trap: Students forget to check that \(f(3)\) also equals 11 — here \(f(3) = 4(3)-1 = 11\). ✓

Apply chain rule: outer function is \(e^u\), inner is \(u = 3x^2-1\).

• Outer derivative: \(e^u\) (stays the same)
• Inner derivative: \(u' = 6x\)
• Result: \(e^{3x^2-1} \cdot 6x = 6xe^{3x^2-1}\)

Common mistake: Writing just \(e^{3x^2-1}\) — forgetting the inner derivative \(6x\)!

Differentiate both sides with respect to \(x\):

• Left: \(3x^2 + 3y^2\dfrac{dy}{dx}\)
• Right (product rule on \(6xy\)): \(6y + 6x\dfrac{dy}{dx}\)
• Collect \(\dfrac{dy}{dx}\): \(3y^2\dfrac{dy}{dx} - 6x\dfrac{dy}{dx} = 6y - 3x^2\)
• Factor: \(\dfrac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2\)
• Result: \(\dfrac{dy}{dx} = \dfrac{6y - 3x^2}{3y^2 - 6x} = \dfrac{2y - x^2}{y^2 - 2x}\)

Key: Product rule on the right side is the tricky part!

• Average rate: \(\dfrac{f(2)-f(0)}{2-0} = \dfrac{(8-6)-0}{2} = \dfrac{2}{2} = 1\)
• Set \(f'(c) = 1\): \(f'(x) = 3x^2-3\), so \(3c^2-3=1\)
• \(3c^2 = 4 \Rightarrow c^2 = \dfrac{4}{3} \Rightarrow c = \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3}\)
• Check: \(\dfrac{2\sqrt{3}}{3} \approx 1.15 \in (0,2)\) ✓

• Let \(u = x^3+4\), then \(du = 3x^2\,dx\), so \(6x^2\,dx = 2\,du\)
• Rewrite: \(\displaystyle\int \dfrac{2}{u^2}\,du = 2\int u^{-2}\,du\)
• \(= 2 \cdot \dfrac{u^{-1}}{-1} + C = -\dfrac{2}{u} + C\)
• Back-substitute: \(-\dfrac{2}{x^3+4}+C\)

LIATE says: \(\ln x\) is a Logarithm → choose \(u = \ln x\).

• \(u = \ln x \Rightarrow du = \dfrac{1}{x}dx\)
• \(dv = x\,dx \Rightarrow v = \dfrac{x^2}{2}\)
• \(\int x\ln x\,dx = \dfrac{x^2}{2}\ln x - \int \dfrac{x^2}{2} \cdot \dfrac{1}{x}\,dx\)
• \(= \dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C\)

FTC Part 2 + Chain Rule (upper limit is \(x^3\), not just \(x\)):

• Apply FTC: result is \(\sqrt{1+(x^3)^2}\) … then multiply by \(\dfrac{d}{dx}[x^3] = 3x^2\)
• Answer: \(3x^2\sqrt{1+x^6}\)

Trap: Forgetting to multiply by the derivative of the upper limit!

• Disk method: \(V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx\)
• \(= \pi\left[\dfrac{x^2}{2}\right]_0^4 = \pi \cdot \dfrac{16}{2} = 8\pi\)

Note: \((\sqrt{x})^2 = x\) simplifies nicely. Don't overcomplicate!

• Separate: \(y\,dy = x\,dx\)
• Integrate: \(\dfrac{y^2}{2} = \dfrac{x^2}{2} + C\) → \(y^2 = x^2 + K\)
• Apply \(y(0)=3\): \(9 = 0 + K \Rightarrow K = 9\)
• Answer: \(y = \sqrt{x^2+9}\) (positive root since \(y(0)=3 > 0\))

This is a geometric series with first term \(a=4\), common ratio \(r=\dfrac{1}{3}\).

• Since \(|r| = \dfrac{1}{3} < 1\), it converges.
• Sum \(= \dfrac{a}{1-r} = \dfrac{4}{1-\frac{1}{3}} = \dfrac{4}{\frac{2}{3}} = 4 \cdot \dfrac{3}{2} = 6\)

Compute \(L = \lim_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n}\right|\):

• \(\dfrac{a_{n+1}}{a_n} = \dfrac{(n+1)^2}{3^{n+1}} \cdot \dfrac{3^n}{n^2} = \dfrac{(n+1)^2}{3n^2}\)
• \(L = \lim_{n \to \infty} \dfrac{(n+1)^2}{3n^2} = \dfrac{1}{3} < 1\)
• Since \(L < 1\), the series converges by Ratio Test.

Simply substitute \(2x\) in place of \(x\) in the known series:

• \(\sin(2x) = (2x) - \dfrac{(2x)^3}{3!} + \dfrac{(2x)^5}{5!} - \cdots\)
• \(= 2x - \dfrac{8x^3}{6} + \dfrac{32x^5}{120} - \cdots\)
• \(= 2x - \dfrac{4x^3}{3} + \dfrac{4x^5}{15} - \cdots\)

Key trick: Never re-derive — just substitute into known series!

• Ratio test: \(L = \lim\left|\dfrac{(x-2)^{n+1}}{(n+1)\cdot 4^{n+1}} \cdot \dfrac{n \cdot 4^n}{(x-2)^n}\right| = \dfrac{|x-2|}{4}\)
• Converges when \(\dfrac{|x-2|}{4} < 1 \Rightarrow |x-2| < 4 \Rightarrow -2 < x < 6\)
• \(x = -2\): \(\sum \dfrac{(-4)^n}{n \cdot 4^n} = \sum \dfrac{(-1)^n}{n}\) → Alternating, converges ✓
• \(x = 6\): \(\sum \dfrac{4^n}{n \cdot 4^n} = \sum \dfrac{1}{n}\) → Harmonic, diverges ✗
• IOC: \([-2, 6)\)

• \(\dfrac{dx}{dt} = 2t\), \(\quad \dfrac{dy}{dt} = 6t^2-3\)
• \(\dfrac{dy}{dx} = \dfrac{6t^2-3}{2t}\)
• At \(t=2\): \(\dfrac{6(4)-3}{2(2)} = \dfrac{24-3}{4} = \dfrac{21}{4}\)

One petal of \(r = 3\sin(2\theta)\) exists from \(\theta = 0\) to \(\theta = \dfrac{\pi}{2}\):

• \(A = \dfrac{1}{2}\int_0^{\pi/2} [3\sin(2\theta)]^2\,d\theta = \dfrac{9}{2}\int_0^{\pi/2}\sin^2(2\theta)\,d\theta\)
• Use identity: \(\sin^2(2\theta) = \dfrac{1-\cos(4\theta)}{2}\)
• \(= \dfrac{9}{2} \cdot \dfrac{1}{2}\int_0^{\pi/2}(1-\cos(4\theta))\,d\theta = \dfrac{9}{4}\left[\theta - \dfrac{\sin(4\theta)}{4}\right]_0^{\pi/2}\)
• \(= \dfrac{9}{4} \cdot \dfrac{\pi}{2} = \dfrac{9\pi}{8}\)

• \(\int_1^{\infty} x^{-1/2}\,dx = \lim_{b\to\infty}\left[2x^{1/2}\right]_1^b = \lim_{b\to\infty}(2\sqrt{b}-2)\)
• As \(b \to \infty\), \(2\sqrt{b} \to \infty\)
• The integral diverges.

p-series connection: \(\int_1^{\infty}\dfrac{1}{x^p}\,dx\) converges only if \(p > 1\). Here \(p = \tfrac{1}{2} < 1\) → diverges.

The next term is degree 4. For \(\cos x\), all derivatives are bounded by 1 in absolute value.

• Error bound: \(|R_3| \leq \dfrac{M \cdot |x|^4}{4!} = \dfrac{1 \cdot (0.1)^4}{24}\)
• \(= \dfrac{0.0001}{24} \approx 4.17 \times 10^{-6}\)

Note: \(\cos x\) Taylor polynomial centered at 0 only has even-degree terms, so \(P_3 = P_2 = 1 - \dfrac{x^2}{2!}\). The next term is \(\dfrac{x^4}{4!}\).

• \(\dfrac{dx}{dt} = -\sin t\), \(\quad \dfrac{dy}{dt} = \cos t\)
• \(L = \int_0^{\pi/2}\sqrt{(-\sin t)^2 + (\cos t)^2}\,dt = \int_0^{\pi/2}\sqrt{\sin^2 t + \cos^2 t}\,dt\)
• \(= \int_0^{\pi/2} 1\,dt = \dfrac{\pi}{2}\)

Geometric check: This is a quarter circle of radius 1, whose arc length is \(\dfrac{1}{4}(2\pi \cdot 1) = \dfrac{\pi}{2}\). ✓

We need a series that is conditionally convergent — alternating series converges, but the series of absolute values diverges.

• \(\displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n}\): alternating harmonic series
• Alternating: ✓ (decreasing, \(1/n \to 0\)) → converges by AST
• Absolute: \(\sum \dfrac{1}{n}\) = harmonic series → diverges
• Therefore it is conditionally convergent

Key distinction: Absolute convergence ≠ Conditional convergence.